Question #15909

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Question #15909

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Answer 1 · 2017-10-25T13:32:39

The hybridisation of $P$ $P$ IN $P C l_{5}$ $PCl_5$ is $s p^{3} d$ $sp^3d$ and the geometry of the compound is Trigonal bi-pyramidal

Explanation:

Hybridisation $=$ $=$ Number of $σ$ $sigma$ bonds $+$ $+$ Number of lone pairs.

In the case of $P C l_{5}$ $PCl_5$ , $P$ $P$ has $5$ $5$ $σ$ $sigma$ bonds and $0$ $0$ lone pairs.

So, the hybridisation is $5$ $5$ .

Therefore it is $s p^{3} d$ $sp^3d$ hybridized.

Now compounds with $s p^{3} d$ $sp^3d$ hybridisation have Trigonal bi-pyramidal geometry.

There are $3$ $3$ $C l$ $Cl$ bonds in the equitorial plane at an angle of $120$ $120$ degrees from each other and $2$ $2$ $C l$ $Cl$ bonds in the axial plane at an angle of $90$ $90$ degrees from the equitorial plane.

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Question #15909

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